Hi! For one of my homework assignments in a math reasoning course I'm asked to prove the following:
If A intersect B=empty set then P(B)-empty set is a subset of P(A union B)-P(A)
I think the power sets may be what's confusing me a bit, so any advice on how to get started would be appreciated.
So far I've tried the "trace an element" method and here's what I have: Choosing an element in P(B)-empty we find that this element is not a member of P(A) because A intersect B=empty. Since P(A union B)-P(A) is P(B), we see that an element of P(B)-empty is also an element of P(A union B)-P(A). Hence, P(B)-empty is a subset of P(A union B)-P(B).
Advice on where to go from here or how to clean this up would be greatly appreciated!!!
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If A intersect B=empty set then P(B)-empty set is a subset of P(A union B)-P(A)
I think the power sets may be what's confusing me a bit, so any advice on how to get started would be appreciated.
So far I've tried the "trace an element" method and here's what I have: Choosing an element in P(B)-empty we find that this element is not a member of P(A) because A intersect B=empty. Since P(A union B)-P(A) is P(B), we see that an element of P(B)-empty is also an element of P(A union B)-P(A). Hence, P(B)-empty is a subset of P(A union B)-P(B).
Advice on where to go from here or how to clean this up would be greatly appreciated!!!
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P(B)-empty set should really read P(B)-{empty set}, for the statement to be true.
The power set of a set S is the set of all subsets of S. So "X is an element of P(S)" means that X is a subset of S. Note that the elements of a power set are themselves sets!
Note also that A and B have no elements in common, since A intersect B is empty.
1) Your statement "choosing an element in P(B)-{empty set} we find that this element is not a member of P(A) because A intersect B=empty" is correct but could use more explanation.
For any element (say X) in P(B)-{empty set}, X is a nonempty subset of B. So X contains at least one element b in B, and b is not in A since A and B have no elements in common. Thus X contains at least one element not in A and so X is not a subset of A, meaning that X is not in P(A).
2) Your statement P(A union B) - P(A) = P(B) is not valid. Consider a set consisting of one or more elements of A, one or more elements in B, but no elements outside of (A union B). This set would be a subset of (A union B) and not a subset of A, but would not be a subset of B.
Remember: the goal is to show that X is not in P(A) and X is in P(A union B).
We have already discussed the proof that X is not in P(A).
To show that X is in P(A union B), recall that X is a subset of B. So all elements of X are in B, and clearly all elements of B are in (A union B). Thus all elements of X are in (A union B). So X is a subset of (A union B), meaning that X is in P(A union B).
Since X is not in P(A) and X is in P(A union B), X is in P(A union B)-P(A).
So X is in P(B)-{empty set} implies that X is in P(A union B)-P(A), and therefore we conclude that P(B)-{empty set} is a subset of P(A union B)-P(A). This completes the proof.
(So, in more concrete language, we have just proved that if A and B have no elements in common, then every nonempty subset of B is a subset of (A union B) and not a subset of A.)
Lord bless you today!
The power set of a set S is the set of all subsets of S. So "X is an element of P(S)" means that X is a subset of S. Note that the elements of a power set are themselves sets!
Note also that A and B have no elements in common, since A intersect B is empty.
1) Your statement "choosing an element in P(B)-{empty set} we find that this element is not a member of P(A) because A intersect B=empty" is correct but could use more explanation.
For any element (say X) in P(B)-{empty set}, X is a nonempty subset of B. So X contains at least one element b in B, and b is not in A since A and B have no elements in common. Thus X contains at least one element not in A and so X is not a subset of A, meaning that X is not in P(A).
2) Your statement P(A union B) - P(A) = P(B) is not valid. Consider a set consisting of one or more elements of A, one or more elements in B, but no elements outside of (A union B). This set would be a subset of (A union B) and not a subset of A, but would not be a subset of B.
Remember: the goal is to show that X is not in P(A) and X is in P(A union B).
We have already discussed the proof that X is not in P(A).
To show that X is in P(A union B), recall that X is a subset of B. So all elements of X are in B, and clearly all elements of B are in (A union B). Thus all elements of X are in (A union B). So X is a subset of (A union B), meaning that X is in P(A union B).
Since X is not in P(A) and X is in P(A union B), X is in P(A union B)-P(A).
So X is in P(B)-{empty set} implies that X is in P(A union B)-P(A), and therefore we conclude that P(B)-{empty set} is a subset of P(A union B)-P(A). This completes the proof.
(So, in more concrete language, we have just proved that if A and B have no elements in common, then every nonempty subset of B is a subset of (A union B) and not a subset of A.)
Lord bless you today!